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$1/0=undefined$, and undefined just means what it says; it's not defined. What happens if we "treat undefined like a thing"? Does that even make sense? I am interested in this since we seem to assign "undefined" to anything whose answer is not defined (correct me if I'm wrong), but can we do anything more with it? I know the tag isn't right, but I don't know what tag would fit, so change it if you can think of a better one.

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    $\begingroup$ It is undefined for a reason-- roughly speaking, treating it like a "thing" on which you can apply standard operations will result in inconsistencies, and wreak havoc. $\endgroup$ – Clement C. Dec 14 '15 at 0:12
  • $\begingroup$ I would say it depends on the definition and the rules of the system you are working in. You can treat it as forking if so DON'T do the normal but do this instead. We sort of do this when we say $x^2/x = x; x\ne 0$. The 0/0 is a flag. Sort of. $\endgroup$ – fleablood Dec 14 '15 at 3:55
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    $\begingroup$ en.m.wikipedia.org/wiki/Wheel_theory $\endgroup$ – Akiva Weinberger Dec 14 '15 at 4:37
  • $\begingroup$ @Clement C: It is completely benign to treat $1/0$ as denoting any number you choose, e.g., $0$. This results in no inconsistencies and no havoc. It just means you have to be careful and check that $z \neq 0$ when you infer $xz = y$ from $x = y/z$, which, of course, you have to do however you approach the issue of undefined terms. $\endgroup$ – Rob Arthan Dec 14 '15 at 22:56
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    $\begingroup$ @RobArthan I don't really understand your point. I was mentioning that assigning a fixed value to $1/0$ on which one can apply the standard and usual rules (addition, multiplication, etc. — and expect them to behave as "expected") would break something. You answer by saying one can very well assign a value —you suggest $0$ — as long as one is careful and checks things before applying the operations and rules. Why not: this may be a good thing to do in some cases; but this is not contradicting what I said. $\endgroup$ – Clement C. Dec 14 '15 at 23:36
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It is possible to use values such as ERROR, UNDEFINED, N/A or MISSING, and give rules for propagating them through calculations. However, this is more useful as a construct in computer programs than for doing or communicating mathematics between people.

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    $\begingroup$ Congratulations for providing a useful answer that is consistent with standard practice in logic and model theory. $\endgroup$ – Rob Arthan Dec 14 '15 at 21:08
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Your question is actually a very interesting one; while it is admittedly impossible to assign a real number to the expression 1/0 since it would prove 1=0, it turns out that we can do something close.

Suppose that, instead of $0$, we take a "real number" which is arbitrarily close to it (called an infinitesimal)—a "real number" which is positive, and yet which is smaller than any actual positive real number. Let us call it $\varepsilon$; we can easily deduce that $\varepsilon$ must have the property that $1/\varepsilon$ is greater than any actual positive real number. Hence, if arithmetic is to have any meaning in our new system, we must also introduce a new "real number", $1/\varepsilon$ which is greater than any actual real numbers.

Further reflection shows that, by multiplying real numbers by $\varepsilon$, we'd also need such things as $2\varepsilon,3\varepsilon,\dots$, and also $\varepsilon/2,\varepsilon/3,\dots$, and thus a "copy" of all the positive numbers, all of which would have the same property that they are somehow sandwiched between 0 and all real numbers—and that's not all! If we consider addition and negation as well, we must conclude that we need to have a "copy" of all real numbers in the "neighborhood" of every real number.

Now, one may of course object to all this construction on the basis that we are still not actually dividing by $0$, but merely a number which we willed to be arbitrarily small. It turns out, however, that considering a thing such as $\varepsilon$ has the convenient property that we can simply take it when we need it and turn it into $0$ again when it is not needed. Such a construction is in fact the underpinning of calculus. In traditional treatments of calculus, the role of having something like $\varepsilon$ is played by limits: consider the definition of the derivative:

$$\frac{df}{dx}=\lim_{\varepsilon\to0}\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$

We can see that "taking the limit" is the operation which makes $\varepsilon$ behave like this infinitesimal number, which we can pretend to not be $0$ when we deal with the division and afterwards make it go away as $0$ again. It turns out, however, that with more advanced techniques one can do away with the limit operation altogether and introduce $\varepsilon$ and its kin directly into the real numbers. An extension of the real numbers in which $\varepsilon$ is an element is called a model of nonstandard analysis; we want to make such an extension elementary (meaning that it satisfies exactly those first-order logical statements which the canonical model $\mathbb{R}$ of actual real numbers satisfies) in order that it also conform to our ideas of how the real numbers ought to behave. It turns out that we can do this by using the Compactness Theorem; an example of this treatment can be found here. More sophisticatedly, one can obtain a model of nonstandard analysis algebraically by taking an ultrapower $\mathbb{R}^\omega/\mathscr{U}$. (Such a construction would admittedly still be non-constructive, given that a non-principal ultrafilter on $\mathbb{R}^\omega$ can only be found by applying an existence lemma.)

Now, given this new structure (call it $\mathbb{R}^*$), even though we still cannot divide by $0$, we now have a proper idea of doing arithmetic with infinitesimals which are essentially $0$. We may show that each nonstandard number (i.e. elements of $\mathbb{R}^*\backslash\mathbb{R}$) is either infinite (i.e. greater than or smaller than all standard real numbers) or is infinitely close to a unique standard real number; this gives us the standard part function $s$ with which we can easily deal with limits. What is traditionally denoted $\lim_{x\to a^+}f(x)=y$ is then replaced by the statement $s(f(a+\varepsilon))=y$, where $\epsilon$ is some infinitesimal.

Coming back to the original question of dealing with such expressions as $1/0$; we may now see that, once we have a model of nonstandard analysis, we can easily just take an infinitesimal $\varepsilon$, and $1/\varepsilon$ behaves exactly as it should be in our model: it is greater than all standard real numbers, though it is still not the greatest: we can yield an even greater infinite number by taking $1/\varepsilon'$ for some "even more infinitesimal" $\varepsilon'$ (i.e. such that $0<\varepsilon'<\varepsilon$). The problem of trying to assign a value to $0/0$ is also disposed of: the value of $\varepsilon_1/\varepsilon_2$ simply depends on the relative "infinitesimalness" of the two infinitesimals; we may choose them in a way that yield any number in return, which corresponds to our intuition of $0/0$. Such a construction is probably the closest you can get to formalizing the idea of "treating division by 0 as a thing" and making the result conform to intuition.

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    $\begingroup$ The ratio 1/0 remains undefined in analysis with infinitesimals, and it is also undefined when considering only integer division of integers, a context where extension by infinitesimals "close" to zero would not make sense. Your answer essentially says that if there are objects $e$ that resemble $0$ in some way, such as having 0 as limit or standard part, or equaling 0 up to 100 decimal places ----- then $1/e$, if defined, will have something in common with some of the intuitions about what 1/0 could be. But if this system with 0-like objects still has 0, then the question of 1/0 persists. $\endgroup$ – zyx Dec 14 '15 at 4:33
  • $\begingroup$ @zyx — I did emphasize the fact in my original answer that dividing by 0 is still impossible—indeed, modulo $\operatorname{Th}(\mathbb{R})$, the sentence $\exists x\,1=x\cdot0$ will always imply $1=0$. However, this doesn't mean that we can't address the problem partially with infinitesimals. Indeed, the very reason that they were introduced was that the early inventors of calculus had to deal with the problem of apparently dividing by zero in e.g. definition of the derivative; hence I believe that a discussion of infinitesimals and extensions to the real numbers that contain them is on topic. $\endgroup$ – Marc Dec 14 '15 at 5:52
  • $\begingroup$ It is standard in first-order logic for function symbols to denote total functions. In model theory it is common to define $1/0 = 0$. This does not lead to a contradiction. It is also standard in computer science to augment domains with a value to denote the result of undefined terms (as in zyx's answer). $\endgroup$ – Rob Arthan Dec 14 '15 at 20:58
  • $\begingroup$ @RobArthan I'm assuming here that / is already defined (as a relational symbol, since it won't be a total function) as a definitional expansion to $\mathbb{R},+,\cdot,0,1$ given by the formula $/(x,y,z)\leftrightarrow x=y\cdot z$, which is what it intuitively should satisfy. $\endgroup$ – Marc Dec 14 '15 at 21:36
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    $\begingroup$ @RobArthan One is of course entitled to define function symbols in an arbitrary expansion as one wishes without leading to contradictions; even defining $x/y=x+y$ would not lead to a contradiction per se. However, since the author of the question asked whether one can hope to make something like $1/0$ "useful", I interpreted this as giving a finer nuance to expressions of the type $1/0$ beyond just giving it a default value contrary to intuition or just propagating the undefinedness of computations. $\endgroup$ – Marc Dec 14 '15 at 21:46
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Suppose that there is a real number (call it $t$) such that $$t = \frac{1}{0}\ .$$ This implies that $1 = t \cdot 0\ .$ Since $t$ is a real number and multiplication with zero yields $0$ we obtain that $1 = 0\ .$ But then $2 = 1+1 = 1+0 = 1\ ,$ and $3 = 2 = 1$ and so on. Consequently, if $\frac{1}{0}$ is a real number then there are no natural numbers, and hence there is no mathematics (since the natural numbers can be thought of as the basis for mathematics).

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  • $\begingroup$ This doesn't answer the question. OP is asking about assigning some kind of 'value' to the expression $\frac 1 0$, and this may very well be something different from a real number. $\endgroup$ – YoTengoUnLCD Dec 14 '15 at 3:52
  • $\begingroup$ Or $1/0$ could be a real number: the law that $x = y/z$ implies $xz = y$ is conditional upon $z \neq 0$. $\endgroup$ – Rob Arthan Dec 14 '15 at 21:06
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    $\begingroup$ Whatever 1/0 may be, it is not a real number (as my post demonstrates). Neither is it a hyperreal number (which the popular post shows). In nonstandard analysis each real number is surrounded by a halo of hyperreal numbers which differ from the real number by an infinitesimal amount. As I see it, the popular post does not answer the OP either, since one cannot assign meaning to 1/0 using nonstandard analysis. $\endgroup$ – Anders Muszta Dec 15 '15 at 11:52

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